Four “4”s

Here is something for U to think over…

Can U get 64 using only two 4’s??

For Example :-

64 using Four 4’s
(4 + 4) x (4 + 4) = 64

64 using Three 4’s
(4 x 4 x 4) = 64

64 using Two 4’s
(    ?  ?    ) = 64

Can U get the number 64 using ONLY TWO 4’s??…

NOTE: ONLY Mathematical operations permitted are
Addition, Subtraction, Multiplication, Division, Exponentiation.

__________________________

Solution(s) : Checkout the COMMENTS.

Meanwhile Do checkout the Computer Nerd’s Acronyms Quiz !!

UPDATE :
People who solved the Puzzle.

Checkout the Hall-of-FAME.


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~ by CVS268 on Mon, 08 Mar, 2010.

24 Responses to “Four “4”s”

  1. 4^4=64. obv from four 4’s =)

    • @ZogG 4^4 = 4*4*4*4 = 256 (NOT 64)

      Good Try!!

      HINT:
      Exponentiation includes squareroot too.
      Right??.. C’mon lets see if U can get 64 now… 😉

  2. 4^(4-4/4)

  3. 4e2*4=64…… (4)^2*4=64

    • U mean Square(4)* 4?…

      64 it is!!

      Good One!! 🙂

      • Come on dude.. where does the 2 come in between in exponent? U need to use two 4’s ONLY!! And Square() is not a usual mathematical function.. If so, I have another most complicated one.. take this..

        4^ceil(Inverse(0.4)) 😉

  4. Got the answer!!! Its..

    4 << 4 😉

    • Woooo Hooooooo!!!!

      Most Innovative answer man!!
      Proud to be an Electronics Engineer!! 😉

      Awesome!! But still one other solution remaining undiscovered…

      • For the uninitiated can you please explain this function/operation that is used in this solution
        4 < < 4 ?

        Thanks!

      • The “<<" is a left shift. In binary i.e. base2 notation, a left-shift by one position is equivalent to multiplying by 2. Hence 4 << 4 is 4 left shifted four times, i.e. doubled four times, which is 64.

  5. Take: 4^4!

    4^4! = 4^24

    Take sqrt(4^24):
    =sqrt(4*4*4*……*4) [sqrt of 24 times 4].
    =2*2*2*……*2 [24 times 2]

    Again take sqrt(2*2*…*2)
    =sqrt( (2*2) * (2*2) * (2*2) * ……) [sqrt of 12 pairs of 2]
    =2*2*2*….*2 [12 times 2]

    Again take sqrt(2*2*…*2)
    =sqrt( (2*2) * (2*2) * ….) [sqrt of 6 pairs of 2]
    =2*2*2*2*2*2 [6 times 2]

    =64

    So the solution is:
    sqrt(sqrt(sqrt(4^4!))).

    • CODE is CRACKED. 🙂

      This is the original solution i had in mind.
      (But the other solutions here are so much more elegant)

      Congrats Rajeev!! 🙂

  6. 44 (in base 15).

  7. 64 = Re( (4^ [ pi ] ) + 4 i)

    where-
    Re = Real part of.
    [x] = Integer function (greatest integer less than x).

    i.e.
    Re( (4^ [ pi ] ) + 4 i)
    = Re(4^3 + 4i)
    = Re(64 + 4i)
    = 64

    (Q.E.D)

  8. answer 1—> 4 ^ -[tan 4!]
    another—-> there u go..
    4 ^ log (sqrt(sqrt(sqrt exp 4!)))

  9. 4^(sqroot(4))

  10. make use of five ZERO’s and make 120. use any mathematical operations…

  11. 4^2×4

  12. 4!!x4!!.
    Duh

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